Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public class C {
	public static void main(String[] args)throws Throwable {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out);
		
		int n=sc.nextInt();
		int m=sc.nextInt();
		int [] minR=new int [n];
		Arrays.fill(minR, 1000);
		int [] minC=new int [m];
		Arrays.fill(minC, 1000);
		int [][] a=new int [n][m];
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++){
				a[i][j]=sc.nextInt();
				minR[i]=Math.min(minR[i], a[i][j]);
				minC[j]=Math.min(minC[j], a[i][j]);
			}
		int [] rows=new int [n];
		int [] columns=new int [m];
		boolean f=true;
		int cnt=0;
		if(m>n){
			for(int i=0;i<n;i++){
				rows[i]+=minR[i];
				cnt+=rows[i];
				for(int j=0;j<m;j++)
					a[i][j]-=minR[i];
			}
			for(int j=0;j<m;j++)
				for(int i=1;i<n;i++)
					if(a[i][j]!=a[i-1][j])
						f=false;
			for(int j=0;j<m;j++){
				columns[j]+=a[0][j];
				cnt+=a[0][j];
			}
		}else{
			for(int j=0;j<m;j++){
				columns[j]+=minC[j];
				cnt+=columns[j];
				for(int i=0;i<n;i++)
					a[i][j]-=minC[j];
			}
			for(int i=0;i<n;i++)
				for(int j=1;j<m;j++)
					if(a[i][j]!=a[i][j-1])
						f=false;
			for(int i=0;i<n;i++){
				rows[i]+=a[i][0];
				cnt+=a[i][0];
			}
		}
		if(!f)
			pw.println(-1);
		else{
			pw.println(cnt);
			for(int i=0;i<n;i++)
				while(rows[i]-->0)
					pw.println("row "+(i+1));

			for(int i=0;i<m;i++)
				while(columns[i]-->0)
					pw.println("col "+(i+1));
		}
		pw.flush();
		pw.close();
	}
	
	
	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			try {
				br = new BufferedReader(new FileReader(new File("car.in")));
			} catch (FileNotFoundException e) {
				br = new BufferedReader(new InputStreamReader(System.in));
			}
			}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
	}
}